CE 374 K – Hydrology Runoff Processes Daene C. McKinney Watershed Watershed Area draining to a stream Streamflow generated by water entering surface channels Affected by Physical, vegetative, and climatic features Geologic considerations Stream Patterns Dry periods Flow sustained from groundwater (baseflow) http://www.epa.gov/owow/watershed/whatis.html Streamflow Atmospheric Moisture Atmospheric Water Rain Evapotranspiration Precipitation Subsurface Water Snow Infiltration Groundwater Surface Water Evaporation Interception Energy Throughfall and Stem Flow Snowpack Snowmelt Pervious Surface Infiltration Soil Moisture Percolation Groundwater Evapotranspiration Overland Flow Groundwater Flow Streams and Lakes Channel Flow Runoff Watershed Boundary Impervious Evaporation Streamflow Hydrograph Basin Lag Centroid of Precipitation Peak Discharge, Q Time of Rise Inflection Point Baseflow Recession Baseflow Recession Beginning of Direct Runoff End of Direct Runoff Time Baseflow Separation No inflow added to groundwater - depletion (recession) curve Continuity equation dS I (t ) Q(t ) dt dS Q0e (t t0 ) / k dt Q(t ) Q0e (t t0 ) / k Discharge, Q Baseflow Recession S (t ) kQ(t ) Q(t ) flow at time t Q0 flow at time t 0 k decay constant [T ] Time Baseflow Separation Techniques Straight – line method Draw a horizontal line segment (A-B) from beginning of runoff to intersection with recession curve Discharge, Q Direct Runoff B A Baseflow Time Baseflow Separation Techniques Fixed Base Method Draw line segment (A – C) from baseflow recession to a point directly below the hydrograph peak Draw line segment (C-D) connecting a point N time periods after the peak N A0.2 Discharge, Q Direct Runoff A C Baseflow D B Time Baseflow Separation Techniques Variable Slope Method Draw line segment (A-C) forward from baseflow recession to a point directly below the hydrograph peak Draw line segment (B-E) backward from baseflow recession to a point directly below the inflection point Draw line segment (C-E) Discharge, Q A Direct Runoff E C Baseflow B Time Abstraction (Losses) Estimation Phi – Index Method Excess (effective) rainfall Abstraction (losses) Rainfall that is not retained or infiltrated Becomes direct runoff Excess rainfall hyetograph (excess rainfall vs time) Difference between total and excess rainfall hyetographs Phi – Index Constant rate of abstraction yielding excess rainfall hyetograph with depth equal to depth of direct runoff M rd Rm t m 1 rd depth of direct runoff Rm observed rainfall Phi index M # intervals of rainfall contributing to driect runoff t time interval Example Have precipitation and streamflow data, need to estimate losses Observed Rain Flow in cfs 8:30 12000 203 9:00 0.15 246 9:30 0.26 283 10:00 1.33 828 10:30 2.2 2323 11:00 0.2 5697 11:30 0.09 9531 12:00 11025 12:30 8234 1:00 4321 0 0.5 10000 1 1.5 2 Streamflow (cfs) Time 8000 2.5 6000 4000 2000 1:30 2246 2:00 1802 2:30 1230 3:00 713 3:30 394 4:00 354 4:30 303 0 7:30 PM 9:00 PM 10:30 PM 12:00 AM 1:30 AM 3:00 AM Time No direct runoff until after 9:30 And little precip after 11:00 Basin area A = 7.03 mi2 4:30 AM 6:00 AM Example (Cont.) Estimate baseflow (straight line method) Constant = 400 cfs 12000 10000 Streamflow (cfs) 8000 6000 4000 2000 0 7:30 PM 9:00 PM 10:30 PM 12:00 AM Time 1:30 AM 3:00 AM 4:30 AM baseflow 6:00 AM Example (Cont.) Calculate Direct Runoff Hydrograph Subtract 400 cfs Time 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 Observed Rain Flow in cfs 0.15 203 0.26 246 1.33 283 2.2 828 2.08 2323 0.2 5697 0.09 9531 11025 8234 4321 2246 1802 1230 713 394 354 303 Direct Runoff cfs 428 1923 5297 9131 10625 7834 3921 1846 1402 830 313 43550 Total = 43,550 cfs Example (Cont.) Compute volume of direct runoff 11 11 n 1 n 1 Vd Qn t t Qn 3600s/hr * 0.5 hr * 43,550 ft 3 /s 7.839*107 ft 3 Compute depth of direct runoff V rd d A 7.839*107 ft 3 7.03 mi * 52802 ft 2 0.4 ft 4.80 in Example (Cont.) Neglect all precipitation intervals that occur before the onset of direct runoff (before 9:30) Select Rm as the precipitation values in the 1.5 hour period from 10:00 – 11:30 M rd Rm t m 1 4.80 (1.33 2.20 2.08 * 3 * 0.5) 0.54 in t 0.27 in rd 4.80 in Example (Cont.) 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 Observed Rain Flow in cfs 0.15 203 0.26 246 1.33 283 2.2 828 2.08 2323 0.2 5697 0.09 9531 11025 8234 4321 2246 1802 1230 713 394 354 303 Direct Runoff Excess Rainfall cfs in 12000 0 t=0.27 0.5 428 1923 5297 9131 10625 7834 3921 1846 1402 830 313 43550 10000 1.06 1.93 1.81 1 1.5 2 Streamflow (cfs) Time 8000 2.5 6000 4000 2000 0 7:30 PM 9:00 PM 10:30 PM 12:00 AM Time 1:30 AM 3:00 AM 4:30 AM 6:00 AM SCS Curve Number Method Soil Conservation Service(SCS) Curve Number (CN) model estimates precipitation excess as a function of cumulative precipitation, soil cover, land use, and antecedent moisture SCS developed the method for small basins (< 400 sq. mi.) to "before" and "after" hydrologic response from events. Classify soils (60 or 70 types) into four hydrologic soil groups Method is simple enough to be used by people that have little experience with hydrology. Converts basin storage into something simpler and more manageable (a “curve number” CN) Abstractions – SCS Method In general Pe P After runoff begins Fa S Potential runoff P Pe I a Fa Precipitation Pe P Ia SCS Assumption Fa Pe S P Ia Solve for Rainfall Excess P I a 2 Pe P Ia S Ia Fa tp Time P Total Rainfall Pe Rainfall Excess I a Initial Abstraction Fa Continuing Abstraction S Potential Maximum Storage SCS Method (Cont.) Experiments showed I a 0.2S So Pe 12 P 0.2S 2 P 0.8S 1000 S 10 CN (American Units; 0 CN 100) 25400 254CN CN (SI Units; 30 CN 100) S Impervious: CN = 100 Natural: CN < 100 100 90 80 70 11 Cumulative Direct Runoff, Pe, in Surface 10 9 60 40 20 10 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 Cumulative Rainfall, P, in 8 9 10 11 12 SCS Method (Cont.) S and CN depend on antecedent rainfall conditions Normal conditions, AMC(II) Dry conditions, AMC(I) CN ( I ) 4.2CN ( II ) 10 0.058CN ( II ) Wet conditions, AMC(III) CN ( III ) 23CN ( II ) 10 0.13CN ( II ) 5-day antecedent rainfall (in) AMC Group Dormant season Growing season I < 0.50 < 1.4 II 0.5 -- 1.1 1.4 – 2.1 III > 1.1 > 2.1 SCS Method (Cont.) SCS Curve Numbers depend on soil conditions Group Minimum Infiltration Rate (in/hr) Soil type A 0.3 – 0.45 High infiltration rates. Deep, well drained sands and gravels B 0.15 – 0.30 Moderate infiltration rates. Moderately deep, moderately well drained soils with moderately coarse textures C 0.05 – 0.15 Slow infiltration rates. Soils with layers, or soils with moderately fine textures D 0.00 – 0.05 Very slow infiltration rates. Clayey soils, high water table, or shallow impervious layer Example - SCS Method - 1 Rainfall: 5 in. Area: 1000-ac Soils: Class B: 50% Class C: 50% Antecedent moisture: AMC(II) Land use Residential Paved roads: 18% with curbs and storm sewers Open land: 16% 40% with 30% impervious cover 12% with 65% impervious cover 50% fair grass cover 50% good grass cover Parking lots, etc.: 14% Example (SCS Method – 1, Cont.) Hydrologic Soil Group B Land use C % CN Product % CN Product Residential (30% imp cover) 20 72 14.40 20 81 16.20 Residential (65% imp cover) 6 85 5.10 6 90 5.40 Roads 9 98 8.82 9 98 8.82 Open land: good cover 4 61 2.44 4 74 2.96 Open land: Fair cover 4 69 2.76 4 79 3.16 Parking lots, etc 7 98 6.86 7 98 6.86 40.38 50 Total 50 CN 40.38 43.40 83.8 43.40 Example (SCS Method – 1 Cont.) Average AMC CN 83.8 S 1000 10 CN 1000 10 1.93 in 83.8 Pe S P 0.2S 2 5 0.2 *1.932 P 0.8S 5 0.8 *1.93 3.25 in Wet AMC CN ( III ) S 23CN ( II ) 23 * 83.8 92.3 10 0.13CN ( II ) 10 0.13 * 83.8 1000 10 0.83 in 92.3 Pe P 0.2S 2 5 0.2 * 0.832 P 0.8S 5 0.8 * 0.83 4.13 in Example (SCS Method – 2) Given P, CN = 80, AMC(II) Find: Cumulative abstractions and excess rainfall hyetograph Time (hr) Cumulative Rainfall (in) P 0 0 1 0.2 2 0.9 3 1.27 4 2.31 5 4.65 6 5.29 7 5.36 Cumulative Abstractions (in) Ia Fa Cumulative Excess Rainfall (in) Pe Excess Rainfall Hyetograph (in) Example (SCS Method – 2) 1000 1000 10 10 2.50 in CN 80 I a 0.2S 0.2 * 2.5 0.5 in Calculate storage Calculate initial abstraction Initial abstraction removes S 0.2 in. in 1st period (all the precip) 0.3 in. in the 2nd period (only part of the precip) Calculate continuing abstraction Pe Fa S P Ia P Pe I a Fa Time (hr) Cumulative Rainfall (in) P 0 0 1 0.2 2 0.9 S (P I a ) 2.5( P 0.5) Fa (P I a S ) ( P 2.0) 3 1.27 4 2.31 5 4.65 2.5(0.9 0.5) Fa (2 hr) 0.34 in (0.9 2.0) 6 5.29 7 5.36 Example (SCS method – 2) Cumulative abstractions can now be calculated Time (hr) Cumulative Rainfall (in) Cumulative Abstractions (in) P Ia Fa 0 0 0 - 1 0.2 0.2 - 2 0.9 0.5 0.34 3 1.27 0.5 0.59 4 2.31 0.5 1.05 5 4.65 0.5 1.56 6 5.29 0.5 1.64 7 5.36 0.5 1.65 2.5( P 0.5) Fa ( P 2 .0 ) Example (SCS method – 2) Cumulative excess rainfall can now be calculated Excess Rainfall Hyetograph can be calculated Time (hr) Cumulative Rainfall (in) P 0 0 1 0.2 2 0.9 3 1.27 4 2.31 5 4.65 6 5.29 7 5.36 R a i nf a l l ( i n) R a i nf a l l H y e t ogr a phs Cumulative Abstractions (in) 2. 5 2 1. 5 1 0. 5 0 Pe P I a Fa Cumulative Excess Rainfall (in) Excess Rainfall Hyetograph (in) Ia Fa Pe 0 - 0 0 0.2 - 0 0 0.5 0.34 0.06 0.06 0.5 0.59 0.18 0.12 0.5 1.05 0.76 0.58 0.5 1.56 2.59 1.83 1.64 3.15 0.56 0.5 0.5 0 1 1.65 2 3 T i me ( h o u r ) 4 3.21 5 0.06 6 7 R a i nf a l l E x c e s s R a i nf a l l Time of Concentration Different areas of a watershed contribute to runoff at different times after precipitation begins Time of concentration Time at which all parts of the watershed begin contributing to the runoff from the basin Time of flow from the farthest point in the watershed